Problem: Which of the following numbers is a multiple of 6? ${85,96,116,117,118}$
Solution: The multiples of $6$ are $6$ $12$ $18$ $24$ ..... In general, any number that leaves no remainder when divided by $6$ is considered a multiple of $6$ We can start by dividing each of our answer choices by $6$ $85 \div 6 = 14\text{ R }1$ $96 \div 6 = 16$ $116 \div 6 = 19\text{ R }2$ $117 \div 6 = 19\text{ R }3$ $118 \div 6 = 19\text{ R }4$ The only answer choice that leaves no remainder after the division is $96$ $ 16$ $6$ $96$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $6$ are contained within the prime factors of $96$ $96 = 2\times2\times2\times2\times2\times3 6 = 2\times3$ Therefore the only multiple of $6$ out of our choices is $96$. We can say that $96$ is divisible by $6$.